The Stacks project

49.15 Comparison with duality theory

In this section we compare the elementary algebraic constructions above with the constructions in the chapter on duality theory for schemes.

Lemma 49.15.1. Let $f : Y \to X$ be a quasi-finite separated morphism of Noetherian schemes. For every pair of affine opens $\mathop{\mathrm{Spec}}(B) = V \subset Y$, $\mathop{\mathrm{Spec}}(A) = U \subset X$ with $f(V) \subset U$ there is an isomorphism

\[ H^0(V, f^!\mathcal{O}_ Y) = \omega _{B/A} \]

where $f^!$ is as in Duality for Schemes, Section 48.16. These isomorphisms are compatible with restriction maps and define a canonical isomorphism $H^0(f^!\mathcal{O}_ X) = \omega _{Y/X}$ with $\omega _{Y/X}$ as in Remark 49.2.11. Similarly, if $f : Y \to X$ is a quasi-finite morphism of schemes of finite type over a Noetherian base $S$ endowed with a dualizing complex $\omega _ S^\bullet $, then $H^0(f_{new}^!\mathcal{O}_ X) = \omega _{Y/X}$.

Proof. By Zariski's main theorem we can choose a factorization $f = f' \circ j$ where $j : Y \to Y'$ is an open immersion and $f' : Y' \to X$ is a finite morphism, see More on Morphisms, Lemma 37.42.3. By our construction in Duality for Schemes, Lemma 48.16.2 we have $f^! = j^* \circ a'$ where $a' : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_{Y'})$ is the right adjoint to $Rf'_*$ of Duality for Schemes, Lemma 48.3.1. By Duality for Schemes, Lemma 48.11.4 we see that $\Phi (a'(\mathcal{O}_ X)) = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (f'_*\mathcal{O}_{Y'}, \mathcal{O}_ X)$ in $D_\mathit{QCoh}^+(f'_*\mathcal{O}_{Y'})$. In particular $a'(\mathcal{O}_ X)$ has vanishing cohomology sheaves in degrees $< 0$. The zeroth cohomology sheaf is determined by the isomorphism

\[ f'_*H^0(a'(\mathcal{O}_ X)) = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(f'_*\mathcal{O}_{Y'}, \mathcal{O}_ X) \]

as $f'_*\mathcal{O}_{Y'}$-modules via the equivalence of Morphisms, Lemma 29.11.6. Writing $(f')^{-1}U = V' = \mathop{\mathrm{Spec}}(B')$, we obtain

\[ H^0(V', a'(\mathcal{O}_ X)) = \mathop{\mathrm{Hom}}\nolimits _ A(B', A). \]

As the zeroth cohomology sheaf of $a'(\mathcal{O}_ X)$ is a quasi-coherent module we find that the restriction to $V$ is given by $\omega _{B/A} = \mathop{\mathrm{Hom}}\nolimits _ A(B', A) \otimes _{B'} B$ as desired.

The statement about restriction maps signifies that the restriction mappings of the quasi-coherent $\mathcal{O}_{Y'}$-module $H^0(a'(\mathcal{O}_ X))$ for opens in $Y'$ agrees with the maps defined in Lemma 49.2.3 for the modules $\omega _{B/A}$ via the isomorphisms given above. This is clear.

Let $f : Y \to X$ be a quasi-finite morphism of schemes of finite type over a Noetherian base $S$ endowed with a dualizing complex $\omega _ S^\bullet $. Consider opens $V \subset Y$ and $U \subset X$ with $f(V) \subset U$ and $V$ and $U$ separated over $S$. Denote $f|_ V : V \to U$ the restriction of $f$. By the discussion above and Duality for Schemes, Lemma 48.20.9 there are canonical isomorphisms

\[ H^0(f_{new}^!\mathcal{O}_ X)|_ V = H^0((f|_ V)^!\mathcal{O}_ U) = \omega _{V/U} = \omega _{Y/X}|_ V \]

We omit the verification that these isomorphisms glue to a global isomorphism $H^0(f_{new}^!\mathcal{O}_ X) \to \omega _{Y/X}$. $\square$

Lemma 49.15.2. Let $f : Y \to X$ be a finite flat morphism of Noetherian schemes. The map

\[ \text{Trace}_ f : f_*\mathcal{O}_ Y \longrightarrow \mathcal{O}_ X \]

of Section 49.3 corresponds to a map $\mathcal{O}_ Y \to f^!\mathcal{O}_ X$ (see proof). Denote $\tau _{Y/X} \in H^0(Y, f^!\mathcal{O}_ X)$ the image of $1$. Via the isomorphism $H^0(f^!\mathcal{O}_ X) = \omega _{X/Y}$ of Lemma 49.15.1 this agrees with the construction in Remark 49.4.7.

Proof. The functor $f^!$ is defined in Duality for Schemes, Section 48.16. Since $f$ is finite (and hence proper), we see that $f^!$ is given by the right adjoint to pushforward for $f$. In Duality for Schemes, Section 48.11 we have made this adjoint explicit. In particular, the object $f^!\mathcal{O}_ X$ consists of a single cohomology sheaf placed in degree $0$ and for this sheaf we have

\[ f_*f^!\mathcal{O}_ X = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(f_*\mathcal{O}_ Y, \mathcal{O}_ X) \]

To see this we use also that $f_*\mathcal{O}_ Y$ is finite locally free as $f$ is a finite flat morphism of Noetherian schemes and hence all higher Ext sheaves are zero. Some details omitted. Thus finally

\[ \text{Trace}_ f \in \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(f_*\mathcal{O}_ Y, \mathcal{O}_ X) = \Gamma (X, f_*f^!\mathcal{O}_ X) = \Gamma (Y, f^!\mathcal{O}_ X) \]

On the other hand, we have $f^!\mathcal{O}_ X = \omega _{Y/X}$ by the identification of Lemma 49.15.1. Thus we now have two elements, namely $\text{Trace}_ f$ and $\tau _{Y/X}$ from Remark 49.4.7 in

\[ \Gamma (Y, f^!\mathcal{O}_ X) = \Gamma (Y, \omega _{Y/X}) \]

and the lemma says these elements are the same.

Let $U = \mathop{\mathrm{Spec}}(A) \subset X$ be an affine open with inverse image $V = \mathop{\mathrm{Spec}}(B) \subset Y$. Since $f$ is finite, we see that $A \to B$ is finite and hence the $\omega _{Y/X}(V) = \mathop{\mathrm{Hom}}\nolimits _ A(B,A)$ by construction and this isomorphism agrees with the identification of $f_*f^!\mathcal{O}_ Y$ with $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(f_*\mathcal{O}_ Y, \mathcal{O}_ X)$ discussed above. Hence the agreement of $\text{Trace}_ f$ and $\tau _{Y/X}$ follows from the fact that $\tau _{B/A} = \text{Trace}_{B/A}$ by Lemma 49.4.3. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0DWM. Beware of the difference between the letter 'O' and the digit '0'.